Question: The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $28.5$ years; the standard deviation is $5.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living less than $39.1$ years.
Explanation: $28.5$ $23.2$ $33.8$ $17.9$ $39.1$ $12.6$ $44.4$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $28.5$ years. We know the standard deviation is $5.3$ years, so one standard deviation below the mean is $23.2$ years and one standard deviation above the mean is $33.8$ years. Two standard deviations below the mean is $17.9$ years and two standard deviations above the mean is $39.1$ years. Three standard deviations below the mean is $12.6$ years and three standard deviations above the mean is $44.4$ years. We are interested in the probability of a zebra living less than $39.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the zebras will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the zebras will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $17.9$ years and the other half $({2.5\%})$ will live longer than $39.1$ years. The probability of a particular zebra living less than $39.1$ years is ${95\%} + {2.5\%}$, or $97.5\%$.